Why is the set $\{e^x, e^{2x}\}$ linearly independent? My (apparently wrong) reasoning that tells me the set is actually linearly dependent goes something like this: 
Take the equation $a_1e^x + a_2e^{2x} = 0$. Assume that $a_1$ doesn't equal 0. Then we have the equation: 
$$\frac{-a_2}{a_1}e^{2x} = e^x \tag{$*$} $$
For all $x$ we can find $a_1, a_2,$ satisfying this equation (we set $a_2 = -1, a_1 = e^x$ for whatever particular $x$ value is being evaluated). 
Is the flaw in my reasoning the fact $a_1$ is dependent on different $x$'s for the equation to work for all $x$'s? If so, to prove that the set was linearly independent, would I just have to remark that there are no constants $a_1, a_2$ that make $(*)$ work for all $x$'s? , and hence contra the assumption, $a_1 = a_2 = 0$? 
 A: Linear dependence doesn't make sense without specifying what the scalars are. If you're allowed to use coefficients that are, say, continuous functions, then $\{ e^x,e^{2x} \}$ is, in fact, linearly dependent, by your very argument: you have a nonzero linear combination
$$ e^x \cdot e^x - 1 \cdot e^{2x} = 0 $$
giving zero.
However, if scalars are restricted to being just real numbers, then the linear combination above doesn't work to show dependence, because $e^x$ is not a scalar.

There is a simplification and an abuse of notation going on here that may be confusing you. Strictly speaking, $e^x$ is a real number (that varies depending on $x$), but the question intends to ask about a function. 
Let me write $f$, $g$, $h$, and $k$ for the four functions defined by
$$ f(x) = e^{2x} \qquad g(x) = e^x \qquad h(x) = 1 \qquad k(x) = 0$$
The question is asking to show that $\{ f, g \}$ is a linearly independent set. I assume we are in the case that scalars are real numbers. So, the question is whether or not there not exist scalars $a,b$, such that
$$af + bg = k $$
(note that $k$ is the zero vector) Now, an equation of functions holds if and only if it holds for all values — so the problem is equivalent to asking of there are scalars $a$ and a $b$ such that, for every $x$, we have
$$ a f(x) + b g(x) = k(x) $$
or equivalently,
$$ a e^{2x} + b e^x = 0 $$
Since you can't find particular scalars $a$ and $b$ that make this equation true for all $x$, the functions are independent.

However, if we take the question literally without recognizing the intended abuse of notation, it is correct to say that the (variable) set of real numbers $\{ e^{2x}, e^x \}$ is linearly dependent (for all values of $x$), by the argument you gave. The thing we need to show has the quantification the other way around: the problem is, for each x, to find an $a$ and a $b$.
A: If $e^x$ and $e^{2x}$ do happen to be linearly independent (which they are) it's impossible to assume that $a_1\neq 0$, since the equation you wrote down would imply that $a_1=a_2=0$. This is where your reasoning fails.
To prove linear independence of (differentiable) functions, we usually want to integrate or differentiate. If we take the first derivative of $a_1e^x+a_2e^{2x}=0$, we get $a_1e^x+2a_2e^{2x}=0$. Subtracting the first equation from the second gives $a_2e^{2x} = 0$, but since $e^{2x}\neq 0$ for all $x$ (in fact, we just need one point where it's nonzero) we see that $a_2=0$. Substituting this then lets us show $a_1=0$ as well.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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If
  $\ds{a\expo{x} + b\expo{2x} = 0\,,\ \color{#f00}{\forall}\ x \in \mathbb{R}}$;
  then it vanishes out , $\ds{\underline{in\ particular}}$ for $\ds{\underline{two}}$ values of $\ds{x}$. For instance: $\ds{x = 0}$ and $\ds{x = 1}$

\begin{align}
&\implies\quad \left.\begin{array}{rcrcl}
\ds{a} & \ds{+} & \ds{b} & \ds{=} & 0 
\\
\ds{\expo{}a} & \ds{+} & \ds{\expo{2}b} & \ds{=} & 0 
\\
\end{array}\right\}\quad\implies
\color{#f00}{a} = \color{#f00}{b} = \color{#f00}{0}
\end{align}

because $\ds{\pars{~1 \times \expo{2} - \expo{} \times 1 \not= 0~}}$.
  
  In another words, $\ds{\not\exists\ \mbox{a constant}\ \lambda\ \mbox{such that}\
\expo{2x} = \lambda\expo{x}\,,\ \forall\ x \in \mathbb{R}}$.

A: If you had a linear combination of $e^x$ and $e^{2x}$ that results in the $0$ function, it would work for all values of $x$. Just pick any two values of $x$ and you'll see that this is impossible.
E.g.: For $x = 0$, $e^x = 1$ and $e^{2x} = 1$. For $x = 1$, $e^x = e$ and $e^{2x} = e^2$. But the vectors $(1,e)$ and $(1, e^2)$ are linearly independent. Q.E.D.
A: To check that $f(x)=e^x$, $g(x)=e^{2x}$ are linearly independent, it is enough to show that for some $x\in\mathbb{R}$
$$\det W(x)=\det\begin{pmatrix}f(x) & g(x) \\ f'(x) & g'(x) \end{pmatrix}\neq 0. $$
For instance, if we consider $x=0$, we have $\det\begin{pmatrix}1 & 1 \\ 1 & 2\end{pmatrix}=1\neq 0$. See also Wronskian.
In a similar way, you may exploit the non-vanishing of a Vandermonde determinant to deduce that $\{e^x,e^{2x},e^{3x},\ldots,e^{Nx}\}$ are linearly independent.
A: The function $e^{2x}= (e^x)^2$ is quadratically dependent on $e^x$, so it is linearly independent of $e^x$. For many decent  functions  it should be true that  $f(x)^2$  and $f(x)$ are linearly independent.
For example assume $f(x)$ is continuously differentiable, and look at the points where the derivative is not zero: from any possible linear dependence $kf(x) =\big(f(x)\big)^2$, for some constant $k\ne0$ we will get, by differentiating, $kf'(x)=2f'(x)f(x)$. And so on the open set where $f'(x)\ne0$  $f$ has to take the constant value $k/2$.
This along with $kf(x) = f(x)^2$ leads to  a contradiction.
